In $0.1 M$ solution, a monobasic acid is $1 %$ ionized. The ionisation constant of the acid is:

1 \times 10^{- 3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 \times 10^{- 7}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 \times 10^{- 5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1 \times 10^{- 14}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $1 \times 10^{- 5}$
Monobasic acid is $1 %$ ionized. The degree of dissociation $α$ is $α = \frac{1}{100} = 0.01$ .

According to Ostwald dilution law equation, $α = \sqrt{\frac{K_{a}}{C}}$ .

Here, $C$ is the concentration.

Thus, $0.01 = \sqrt{\frac{K_{a}}{0.1}}$ or $K_{a} = 10^{- 5}$ .

Option C is correct.

Suggest Corrections

Similar questions

A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is

[H^{+}]

{[O H]}^{-}

[i o n i c p r o d u c t o f w a t e r = 1 \times 10^{- 14}]

3 %

0.1 M

25^{\circ} C

10^{- 14}

0.1 M

2 %

1 \times 10^{- 14}

[O H^{-}]

2.24 \times 10^{- 2}

p H

0.01 M

Join BYJU'S Learning Program